Problem: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{y^3 - 15y^2 + 54y}{9y^3 - 9y^2 - 270y}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {y(y^2 - 15y + 54)} {9y(y^2 - y - 30)} $ $ a = \dfrac{y}{9y} \cdot \dfrac{y^2 - 15y + 54}{y^2 - y - 30} $ Simplify: $ a = \dfrac{1}{9} \cdot \dfrac{y^2 - 15y + 54}{y^2 - y - 30}$ Since we are dividing by $y$ , we must remember that $y \neq 0$ Next factor the numerator and denominator. $ a = \dfrac{1}{9} \cdot \dfrac{(y - 6)(y - 9)}{(y - 6)(y + 5)}$ Assuming $y \neq 6$ , we can cancel the $y - 6$ $ a = \dfrac{1}{9} \cdot \dfrac{y - 9}{y + 5}$ Therefore: $ a = \dfrac{ y - 9 }{ 9(y + 5)}$, $y \neq 6$, $y \neq 0$